A resistor triangle

Electricity and Magnetism Level 4

In the above circuit, all the resistors marked $R$ have a resistance equal to $5 \Omega$ . All of the batteries have electromotive forces of $4.5 \text{V}$ and an internal resistance of $3 \Omega$ .

Calculate the intensities of the currents $I_R$ and $I_B$ through the resistors marked $R$ and through the batteries, respectively. Input the answer as the sum of the intensities of these two currents in Amperes. If the polarities of the currents are negative, reverse them.

The answer is 1.5.

1 solution

Djordje Veljkovic
Jan 25, 2017

The circuit is symmetric and all the three corners of the triangle are identical. This implies points $A$ , $B$ , and $C$ have the same potential and the potential differences $\Delta V$ across all the resistors $R$ are zero.

Now, According to the Ohm's Law $\Delta V = i R.$ As $\Delta V$ across $R$ is zero, therefore, the current through them $I_R$ is also zero.
Now, as the resistors serve no purpose, we can easily calculate the current going through the batteries by applying Kirchoff's Voltage Law $I_B = \frac{3E}{3R_g}$ $I_B = 1.5 A.$ The answer should be $I_R + I_B = I_B = \boxed{1.5 A}.$

Can you explain how is this circuit symmetric

Sahasra Ranjan - 4 years, 1 month ago

Of course! Let's look at some simple geometry first. We know that equilateral triangles have three points of symmetry. To be sure of this, you can draw an triangle, and draw one of its heights. It's easy to notice that the height divides the triangle into two identical parts. What goes for normal geometry, goes for circuits as well, meaning that any circuit with three vertices between which there are elements of the same specifications is symmetric. For a better explanation, with proof, please go to this link: http://www.ittc.ku.edu/~jstiles/723/handouts/Symmetric%20Circuit%20Analysis.pdf

I hope this helps!

Djordje Veljkovic - 4 years, 1 month ago

A resistor triangle

The answer is 1.5.

1 solution

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